Mathematical Foundations - Control of Color Imaging Systems: Analysis and Design - page 108

S OLUTION
The characteristic equation 1 75D 1 125D 2 0 : þ 0 : ¼ 0 can be written as

D 2 0 : 75D þ 0 : 125 ¼ (D 0 : 5)(D 0 : 25) ¼ 0 (3 : 23)

Therefore, the roots are D 1 ¼ 0 : 5 and D 2 ¼ 0 : 25. The zero-input response is

5) n 25) n y(n) ¼ C 1 (0 : þ C 2 (0 : (3 : 24)

The constants C 1 and C 2 are found by applying the initial conditions

5) 1 25) 1 y( 1) ¼ C 1 (0 : þ C 2 (0 : ¼ 2C 1 þ 4C 2 ¼ 1 (3 : 25)

5) 2 25) 2 y( 2) ¼ C 1 (0 : þ C 2 (0 : ¼ 4C 1 þ 16C 2 ¼ 1 (3 : 26)

The solutions are C 1 ¼ 1 : 25 and C 2 ¼ 0 : 375. Therefore,

5) n 25) n

y(n) ¼ 1 : 25(0 : þ 0 : 375(0 : n 2 (3 : 27)

Next, we consider the homogenous solution. The homogenous solution is found
by driving the system with an input but zero initial conditions. The homogenous
solution is similar to input signal; for example, if the input is an exponential function,
the response will be exponential, if the input is a sinusoid, the output is also a sinusoid.

Example 3.9

Consider the following second-order DE:

y(n) ¼ 0 : 75y(n 1) 0 : 125y(n 2) þ u(n) (3 : 28)

4) n Find the output signal y(n) if the input signal is u(n) ¼ 2(0 : for n 0. Assume

zero initial conditions, that is, let y( 2) ¼ y( 1) ¼ 0.

S OLUTION
Since the input is an exponential function, the homogenous solution will be
another exponential, that is,

4) n

y h (n) ¼ A(0 : (3 : 29)

Substituting Equation 3.29 into Equation 3.28 yields

4) n 4) n1 4) n2 4) n A(0 : ¼ 0 : 75A(0 : 0 : 125A(0 : þ 2(0 : (3 : 30)

Simplifying the right side of Equation 3.30, we have

4) n 4) 1 A 4) 2 A þ 4) n A(0 : ¼ [0 : 75(0 : 0 : 125(0 : 2](0 : (3 : 31)